Hot air ballooning

The idea is that the balloon experiences a buoyant force due to the air surrounding it. The air inside the balloon displaces a volume of air outside and hence experiences a buoyant force.

The magnitude of this force is: $F_b = \rho_{out} V g$ where:

$V = \dfrac{4 \pi r^3 } {3}$ is the volume of the balloon's sphere

and

$\rho_{out}$ is the density of the air outside. This is given by $\rho_{out} = \dfrac {P_0 \mu } {R T_0}$ with $P_0$ and $T_0$ being the outside pressure and temperature respectively.

Now, for this balloon to be usable, this must be able to lift the total mass of the balloon. The total mass includes the $300 \text{ kg }$ and also that of the mass of the air inside the balloon. Let's call it $m = \rho_{in} V = \dfrac {P_0 \mu } {R T} V$ . Notice that I used the same pressure term $P_0$ . That's because if it weren't the case, then the balloon will expand until the pressure inside becomes equal to the pressure outside.

Ok. So let's write the god equation: $F_{net} = ma$ . Since here we are asked for the minimum radius, we can say that it is enough if we just balance the buoyant force and the weight force.

$F_b = (300 + m) g$ .

Plugging in all the stuff, and solving for $r$ , the radius, gives the answer as $6.15656$ . Of course, I used Wolfram alpha for the final calculation!

The temperature of the gas varies from the ambient temperature of 20 degrees Celsius (or 293K) to a maximum of 120 degrees Celsius (393K). Applying Charles' Law (V/T=V/T) shows that the total volume for a given mass of the heated air is 393/293 times greater than that of the unheated air. This means that for a container with constant size and pressure (the balloon), heating the gas will result in 100/393 escaping the container (since pressure remains the same).

This fraction of the gas must have a mass equal to 300 kg for the remaining gas in the balloon to have enough buoyant force to remain in the air. 100/393 of the gas has a mass of 300kg, so the total gas volume has a mass of 300*393/100 kg = 1179kg. This is the mass of the volume of unheated gas that will fill the balloon.

1179kg/(0.029kg/mol) = 40655 moles of air in the unheated gas volume. Plugging this into the Ideal Gas Law, we find that PV=nRT (101325 Pa) V = (40655mol)(8.31)(293K) Solving for V gives us V = 976.9 cubic meters Setting that equal to the volume of a sphere (4/3)(pi)(r^3) gives us a radius of 6.15 meters.

The first and most important thing to know in solving this problem is how a hot air balloon operates. The air inside the balloon is hotter and therefore less dense than the air outside the balloon. This causes the less dense, hotter air to create an upward force as it tries to move upward through the more dense, cooler air.

This upward force is equal to the difference in the densities times the volume of hot air, or $F = V_{balloon} \times | \rho_{outside} - \rho_{inside} |$ .

The shape of the balloon is a perfect sphere, so $V_{balloon} = \frac{4}{3} \pi r^3$ .

Substituting into the original formula, this gives $F = \frac{4}{3} \pi r^3 \times | \rho_{outside} - \rho_{inside} |$ .

To solve for the densities, we need to use a variant of the ideal gas law, $P V = n R T$ .

In this equation, $P$ is pressure (in our case, in atmospheres), $V$ is volume (in liters), $n$ is moles of gas, $R$ is the molar gas constant (for our purposes, $.0821 \: \mathrm{\frac{L \cdot atm}{mol \cdot K}}$ ), and $T$ is temperature (in Kelvin).

By multiplying both sides by mass and dividing both sides by $V n R T$ we get $\frac{m}{V} = \frac{mP}{n R T}$ . Substituting for density and molar mass this simplifies to $\rho = \frac{\mu P}{R T}$ .

Plugging in with the temperatures inside the balloon ( $120^\circ \mathrm{C} = 393 \: \mathrm{K}$ ) and outside the ballon ( $20^\circ \mathrm{C} = 293 \: \mathrm{K}$ ) we get densities of $\rho_{inside} = 0.899 \: \mathrm{g / L}$ and $\rho_{outside} = 1.206 \: \mathrm{g / L}$ .

Plugging these results back into the original equation we have:

$F = \frac{4}{3} \pi r^3 \times | 1.206 \: \mathrm{g / L} - 0.899 \: \mathrm{g / L} |$

$= \frac{4}{3} \pi r^3 \times .307 \: \mathrm{g / L}$

$= (1.286 \: \mathrm{g / L}) \times r^3$ .

Finally, the minimum force of lift must be $300 \: \mathrm{kg}$ to overcome the force of gravity on the total mass of the balloon. Substituting and solving gives:

$300 \: \mathrm{kg} = (1.286 \: \mathrm{g / L}) \times r^3$

$233.281 \: \mathrm{m^3} = r^3$ (Note: $\mathrm{g / L} = \mathrm{kg / m^3}$ )

$r = \sqrt[3]{233.281 \: \mathrm{m^3}} = 6.156 \: \mathrm{m}$

F lift = m total * g

V balloon * (d air(293 Kelvin) - d_air(393 Kelvin)) * g = 300 * g

d air = P / (R spesific * T) and densities are;

d air(293 Kelvin) = 1.204 kg m^-3 d air(393 Kelvin) = 0.898 kg m^-3

putting these values in the equation above, we get the volume of the balloon

V_balloon = 4 * pi * r^3 / 3 = 980.4

r = 6.16 m

The minimum radius of our balloon will be determined when the total density of the balloon system is equal to the density of the surrounding air. We can express this as $\dfrac{m + \rho_{hot}V}{V} = \rho_{cool}$ where $\rho_{hot}$ represents the density of the warm air inside the balloon and $\rho_{cool}$ represents the density of the surrounding air. We are given that m = 300 kg and we know that $V = \dfrac{4}{3}\pi r^3$ . We can calculate the air densities using dimensional analysis and the ideal gas constant $R = \dfrac{0.0821\ L\ atm}{mol\ K}$ : $\rho_{cool} = \dfrac{29\ g}{mol} \times \dfrac{mol\ K}{0.0821\ L\ atm} \times \dfrac{1\ atm}{293\ K} \times \dfrac{kg}{1000\ g} = 1.206 \times 10^{-3} \frac{kg}{L}$ Note: 293 K = 20º C. Now, notice that as the temperature increases, the air density decreases, this implies an inversely proportional relationship between density and pressure, that is: $\rho_{cool}T_{cool} = \rho_{hot}T_{hot}$ The warm temperature is 120º C = 393 K. Plugging in our previously determined value into the equation and solving, we obtain $\rho_{hot} = 8.99 \times 10^{-4} \frac{kg}{L}$ Solving the first equation for V, we get $V = \dfrac{m}{\rho_{hot}-\rho_{cool}} = 977198.7\ L$ Now we convert liters into cubic meters: $977198.7\ L \times \dfrac{1000\ mL}{L} \times \dfrac{1 {cm}^3}{1 \ mL} \times \dfrac{1^3\ m^3}{100^3\ {cm}^3} = 977.2\ m^3$ Solving $\dfrac{4}{3}\pi r^3 = 977.2\ m^3$ for r yields 6.15 m, our final answer.